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Suppose we have a qutrit with the state vector $|\psi\rangle = a_0|0\rangle + a_1|1\rangle + a_2|2\rangle$, and we want to project its state onto the subspace having the basis $\{|0\rangle,|2\rangle\}$, I know the projection operator would be written like: $1|0\rangle \langle0| + 0|1\rangle\langle 1| + 1|2\rangle\langle2|$.

I'm having a few confusions here. Does $|0\rangle \langle 0|$ represent a tensor product between $[1 \ 0 \ 0]^{T}$ and $[1 \ 0 \ 0]$ ? Or is it just matrix multiplication? Also, I thought that we must always be able to write a projection operator in the form $|\phi\rangle \langle \phi|$ where $|\phi\rangle$ is a possible state of a qutrit. But how to represent $1|0\rangle \langle0| + 0|1\rangle\langle 1| + 1|2\rangle\langle2|$ in the form $|\phi\rangle \langle \phi|$?

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Does $|0\rangle\langle0|$ represent a tensor product or is it just matrix multiplication?

You can think of $|0\rangle\langle0|$ as tensor product of $|0\rangle$ and $\langle0|$, or equivalently as the matrix multiplication (more precisely, Kronecker product) of the vectors representing them.

Also, I thought that we must always be able to write a projection operator in the form |ϕ⟩⟨ϕ|

Not necessarily. A projector will have that form if and only if it projects onto a one-dimensional space (that is, it projects onto a pure state). More general projections, like the one you mention, do not have this feature, and that is totally fine.

Indeed, also the identity matrix is a (trivial) projection, and it certainly cannot be written as $|\psi\rangle\langle\psi|$ for any pure state $|\psi\rangle$.

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    $\begingroup$ It's the same in this case, but I always thought of $|0\rangle\langle 0|$ as normal matrix product. $\endgroup$ – Ali Jun 10 '18 at 7:28
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A projection operator $P$ has two key properties: $$ P^\dagger=P\qquad P^2=P $$ A particularly simple instance of a projection operator is a rank 1 projector, $P=|\phi\rangle\langle\phi|$, which you can easily see satisfies the two properties given that $|\phi\rangle$ is a normalised state, so $\langle\phi|\phi\rangle=1$.

To see what rank the projector is, simply evaluate $\text{rank}(P)=\text{Tr}(P)$. In your state example of $P=|0\rangle\langle 0|+|2\rangle\langle 2|$, you can see that the rank is 2, so it cannot be written as a rank 1 projector $|\phi\rangle\langle\phi|$.

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